By Harrison D.M.

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**Example text**

The reason noW' is clear. Unfortunately, with these weaker assumptions, the results are not nearly so nice. In Section 3 of these notes we will return to the discussion of Part A(ii) with the weaker hypotheses. E11G = E Now for the sketch. We have {Xl = 1, xZ""'x } n Let 1 Since by be a transversal for be the inverse image in E. g Xi cial. If 1y, (et: is nons i ngular on then El E x. 1 El 1 F [n ]-module. Let is a 1 G so that so that E. so that 1 x. J so that i~ E. 1 also extras pe- [Ei,E ] = 1.

In Section 3 of these notes we will return to the discussion of Part A(ii) with the weaker hypotheses. E11G = E Now for the sketch. We have {Xl = 1, xZ""'x } n Let 1 Since by be a transversal for be the inverse image in E. g Xi cial. If 1y, (et: is nons i ngular on then El E x. 1 El 1 F [n ]-module. Let is a 1 G so that so that E. so that 1 x. J so that i~ E. 1 also extras pe- [Ei,E ] = 1. 2)) is tIle central product of the E 's and G factors just as it permutes the suramands El V lE in E is an extraspecial group.

1 may - 33 - vary from one indecomposable summand U of V to another. 1) Problem. Are irreducibly generated modules algebraic (integral)? For solvable groups, the answer is yes. He will prove this theorem as an appli- cation of the method outlined in Section 1. 3) Proposition. re integral. 7) and proof of a stronger theorem We will first prove the following proposition. u}~~~ V V is algebraic. 3) to fit precise ly into the method. Before we beein the method, we prove enough propositions so that we can move smoothly through the method without interruption..